Conservation of phase space volume in Rindler space-time Let us consider Rindler space-time, i.e. Minkowski space-time as seen by a constantly accelerating observer. My question is, does Liouville's theorem, i.e. the conservation of phase space volume in classical Hamiltonian mechanics, hold in Rindler space-time?
In other words, for an ensemble defined on any arbitrary system of interacting classical particles and fields with a Lorentz-invariant Lagrangian, is the phase space volume at Rindler time $t=t_1$ necessarily equal to the phase space volume at $t=t_2$?  More or less equivalently, if we transform a Hamiltonian system from Minkowski space-time coordinates to Rindler ones, is the result still a Hamiltonian system?
I'm interested in the answer to this for two reasons. The first is that it would immediately provide an answer to another question of mine, Phase space volume and relativity. The other is that if the answer is "yes" then there is no analogue of the black hole information paradox in Rindler space-time (at least classically), whereas if it's "no" then there is, and it might be easier to think about the solution in terms of Rindler space-time than in terms of black holes.
I guess the quantum version of this question would be something like "does unitarity hold for a quantum field theory defined on the Rindler wedge?" I am also interested in this question, although I haven't studied QFT so I might not understand an overly technical answer. I guess the answer to this quantum version is "no", with the lack of unitarity at the horizon being the origin of Unruh radiation -- but I'd like to know if this intuition is correct.
 A: An approach alternative to that discussed by   David Bar Moshe is to start from a different coordinate system in the Rindler wedge $W_R$:
$$ds^2 = e^{2y}(−g^2dt^2+dy^2)$$
here $t, y \in \mathbb R$. The relation with the standard spatial coordinate in $W_R$ is $x=e^y$, where $x>0$ is related with the alternate form of the (same) metric:
$$ds^2 = -g^2 x^2 dt^2 + dx^2\:.$$
In both cases $\partial_t$ is the timelike Killing vector field given by the Lorentzian boost. As in David Bar Moshe's answer the Lagrangian is proportional to the arch length:
$$L(y,\dot{y}) = m e^{2y}\sqrt{g^2 - \dot{y}^2}\:.$$
where $\dot{y}= dy/dt$. So, notice that the relevant notion of time is the Rindlerian one $t$.
It is true that $W_R$ is not geodesically complete and the solutions of Lagrange equations are, in fact, timelike geodesics, but these geodesics need an infinite amount of time to reach the boundary of $W_R$.
So I am confident that dynamics is well defined in $W_R$ and quantization, too.
As already stressed by  David Bar Moshe,  the fact that an Hamiltonian formulation exists (it is enough to perform the standard Legendere transformation) implies that there exists a conserved Liouville measure $\omega =  dq \wedge dp$.
If I am not wrong with my computations, defining $p = \frac{\partial L}{\partial \dot{q}}$ as usual, the Hamiltonian function in Hamiltonian variables turns out to be now:
$$H(y,p) = g \sqrt{p^2 + e^{2y}m^2}\:.$$
This leads to a well-defined quantization procedure as it follows.
(1) The Hilbert space is $L^2(\mathbb R, dy)$, $dy$ being the standard Lebesgue measure (with the choice of the coordinate $x$, the natural space would be defined on $\mathbb R_+$ instead).
(2) The position and momentum operators are $\hat{y}$ multiplicative on it natural dense domain and the unique self-adjoint extension $\hat{p}$ of $-i\frac{d}{dy}$ initially defined on the core $C_0^\infty(\mathbb R)$.
(3) The Hamiltonian operator is the self adjoint opertaor $H= g\sqrt{F}$, where $F$ is Friedrichs' self-adjoint extension of the positive symmetric operator
$$F: \hat{p}^2 + m^2 e^{2\hat{y}} : C_0^\infty \to L^2(\mathbb R, dy)\:.$$
As $F$ is self-adjoint and positive, $\sqrt{F}$ is well defined via spectral theory.
So there is at least one unitary dynamics in Rindler wedge. Also notice that the time parameter of the unitary evolution is naturally identified to Rindler time $t$.
If $\psi$ is sufficiently regular and solves Schroedinger equation $$-i\partial_t \psi = H\psi$$
it also solves $-\partial^2_t\psi = H^2 \psi$, namely
$$-\frac{\partial^2 \psi}{\partial t^2} = -g^2 \frac{\partial^2 \psi}{\partial  y} + g^2 e^{2y} m^2 \psi(t,y)\:.$$
Rearranging it, we find just Klein-Gordon equation in coordinates $t,y$:
$$-g^{-2} e^{-2y} \frac{\partial^2 \psi}{\partial t^2} +  e^{-2y}\frac{\partial^2 \psi}{\partial  y^2} -  m^2 \psi(t,y) =0\:,$$
as expected.
Actually one should study if the naive Hamiltonian operator admits further
self-adjoint extensions than that I pointed out. I expect that one of these extensions has some interplay with another quantization procedure based on conformal group I studied several years ago with a colleague (Nuclear Physics B 647 (2002) 131–152).
Added note.  In principle one may add every interaction to a system of particle living in $W_R$ separately described by corresponding Hamiltonians $H$ as above, without any problem. The only difficulty is that this picture cannot describe the physical system traversing the horizon (as seen within the Minkowskian reference frame), because it happens at the end of time. In this sense the theory is not equivalent to the Minkowskian one, even restricting to the Rindler wedge.
This fact entails that there is no canonical transformation between Minkowskian Hamiltonian formalism and Rindlerian Hamiltonian one as the corresponding "space-times of phases" are different (are associated with  different $(q,p)$-foliations because referred to different notions of time evolution).
So, your two questions are in fact different and have different answers: In Rindler space there is a standard Hamiltonian description, whose time evolution preserves the volume of the relevant phase space. Conversely, there is no canonical transformation between the Minkowskian phase space and the Rindlerian one. They are different spaces of phases and there is no conserved volume.
A: The dynamics of a classical point particle moving in the background of any curved space-time is always Hamiltonian (with respect to the canonical symplectic form), thus automatically satisfying the Liouville’s theorem. This is because the action functional is given by the integral of the line element:
$$ I = -m \int ds = -m \int \frac{ds}{dt}(q, v) dt = \int L(q, v) dt$$.
(Here  $ v = \dot{q}$ are the velocities and $m$ is the mass).
The generalized momenta are the derivatives of the Lagrangian with respect to the velocities:
$$ p = \nabla_{v}L$$
Which is solved for $v$ : $v = v(q, p)$
The Hamiltonian
$$ H(p, q) = pv(q, p) - L(q, v(q, p))$$
Hamilton's equations of motion generate the geodesic motion on the space-time.  This is a sufficient condition for the validity of Liouville's theorem.
In the Rindler case in $(1+1)\mathrm{d}$, we have:
$$ds^2 = -g^2 x^2 dt^2 - dx^2$$
Thus
$$L = -m \sqrt{g^2 x^2 - \dot{x}^2}$$
The generalized momentum
$$ p = \frac{m \dot{x}}{\sqrt{g^2 x^2 - \dot{x}^2}}$$
The Hamiltonian
$$ H = gx \sqrt{p^2+m^2}$$
However, there is a difficulty in the quantization of this Hamiltonian, because its integral curves (the geodesics) reach the boundaries after a finite length. This means that it does not have a self-adjoint quantization. Please see for example, the following post by Terry Tao and also the answer on this question  by Emilio Pisanty on Physics StackExchange.
