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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.

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To ask bluntly: You are using this approach to question certain aspects of the equivalence principle, yes? My reflex reaction was to say that under EP the phase space volumes at $t_1$ and $t_2$ must be equivalent. Or am I just badly misunderstanding your intent? More specifically: I don't see gravity changing phase space volumes over time, so I would not anticipate it for acceleration either. –  Terry Bollinger Feb 24 at 3:45
    
@TerryBollinger I initially thought that the equivalence principle would give an easy answer to this question, but after trying to make it work mathematically I realised (I think) that the equivalence principle is only local, even in flat space-time. I agree that gravity shouldn't change phase space volumes over time, but I don't know how to prove it. Ultimately my main motivation is just that I want to prove that entropy is Lorentz-invariant in classical statistical mechanics. It seems so obvious that it should be (and many have assumed it is), yet an actual proof seems rather elusive. –  Nathaniel Feb 24 at 4:07
    
Intriguing... do you have any papers on your analysis so far? Also, just a random thought: time dilation relative to an unaccelerated observer also changes over time for acceleration. Since "hidden observers" can sneak into equations in surprisingly subtle ways, is it possible you have such a hidden unaccelerated observer implicit in you formalisms? (Even more random: time dilation vs phase space reductions: any connections?) –  Terry Bollinger Feb 24 at 16:38
    
@TerryBollinger no papers I'm afraid - for the moment this is just some very long term idle speculation, and slightly outside my field. If I manage to answer these questions I'll probably publish it though! –  Nathaniel Feb 25 at 8:07

2 Answers 2

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.

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I'd make the same comment on this that I made on David Bar Moshe's answer: it's helpful (and I'll study it in detail), but it only covers the case of a single point particle with no forces. I'm interested in whether the result holds in general, for a system that might have many interacting components and might have a significant spatial extent. –  Nathaniel Feb 27 at 12:17
    
Yes, in principle you may add every interaction to a system of particle separately described by corresponding Hamiltonians $H$ as above, without any problem. The only difficulty is that, this picture cannot describe the system traversing the horizon, 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. –  V. Moretti Feb 27 at 12:43
    
If I made the Hamiltonian up out of several separate Hamiltonians of that form, it would correspond to non-interacting particles. I'm interested in the general case, of several particles that can interact. –  Nathaniel Feb 27 at 15:30
    
You can add interactions, something like, for instance, $H_I(y,y')= kyy'$, with total Hamiltonian $H(y,y',p,p') = g\sqrt{p^2+ e^{2y}m^2} + g\sqrt{p'^2+ e^{2y'}m'^2} + kyy'$. What is the problem with such a Hamiltonian? Maybe I am not understanding your point... –  V. Moretti Feb 27 at 15:33
    
Well, just that if I have a Hamiltonian specified in terms of Minkowski coordinates, I don't know how to transform the interaction terms into that kind of form, or whether it's always possible to do so. In fact, for an arbitrary Hamiltonian it obviously isn't possible, since a particle might be interacting with another particle outside the Rindler wedge. However, if we restrict ourselves to Hamiltonians without superluminal interactions (whatever that means formally) then it might always be possible. That's what I'd like to show. –  Nathaniel Feb 28 at 3:32

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.

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Thanks, this is helpful. However, a single point particle is very much a special case. I'm specifically interested in whether this result holds in the general case, where the system might be composed of several interacting particles and/or fields, and have a non-negligible spatial extent. –  Nathaniel Feb 25 at 15:19
    
@Nathaniel In general, the phase space of a system of several particles, even in interaction, is the Cartesian product of their individual phase spaces. The phase space is the space of initial data and one needs two initial conditions per particle. Thus, the same result should be valid. –  David Bar Moshe Feb 25 at 16:09
    
@Nathaniel cont. In the field case, you are talking about an infinite dimensional Hamiltonian system. It is not straightforward to define a volume in infinite dimensions. (It must be regularized). Although, there are some results on regularized volumes in general, I think that there is only little work, if any, on Liouville's theorem in infinite dimensions. –  David Bar Moshe Feb 25 at 16:10
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@DavidBarMoshe "the phase space of a system of several particles, even in interaction, is the Cartesian product of their individual phase spaces" - I think you might be conflating phase space with the volume of phase space taken up by an ensemble. It's the latter that's conserved (and what I'm interested in), and it's only given by the volume of the Cartesian product if the particles are statistically independent, which in general they won't be. You need a lot more than two initial conditions per particle to specify an ensemble, since you also have to specify correlations. –  Nathaniel Feb 26 at 10:57
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@DavidBarMoshe I only mention fields because relativity says all interactions have to be local, so it's hard to imagine how particles can interact in relativistic classical physics other than via fields. (Or by point collisions I guess, but that feels like a special case.) I'd be happy to avoid thinking about fields if I thought I could get away with it. –  Nathaniel Feb 26 at 11:00

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