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In classical mechanics, by knowing the present, is it always possible to uniquely reconstruct the past? By knowing the phase space point at present i.e., the set of coordinates $\{q_i(0),p_i(0)\}$, for all $i$, is it possible to tell which point the system came from?

If not, then I would like to see why, how maybe with a simple example.

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by knowing the present, is it always possible to uniquely reconstruct the past?

Yes. This is a consequence of Liouville's theorem. According to Liouville's theorem the "volume" in phase space is constant as the system evolves. So if you start with a single point in phase space at one moment in time then at any other moment in time the phase space will also be a single point. It doesn't matter if you trace the phase space evolution forwards or backwards.

This theorem is based on a Hamiltonian with a conserved energy, so it applies at a fundamental level, but may not apply if you are neglecting microscopic degrees of freedom. So “knowing the present” means exact knowledge of the present state including all microscopic degrees of freedom.

Of course, in practice you never know the present exactly. So in reality our practical ability to "predict" the past degrades as we get further away just the same as our ability to predict the future.

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  • $\begingroup$ What happens to the case where the phase space volume shrinks with time (say, in the case of a damped oscillator)? Your answer would suggest that in that case, the past of the oscillator cannot be uniquely constructed even if we knew the phase space point exactly at present? $\endgroup$ Sep 3, 2021 at 14:47
  • $\begingroup$ @mithusengupta123 Yes, of course, dissipative systems do not conserve phase-space volume -- however, it is only because we are choosing to focus on a part of the system and not on the system as a whole. If you were to consider the full phase space of all degrees of freedom in the case of a damped harmonic oscillator, such as the kinetic degrees of freedom of the spring and the environment into which the energy dissipates, you'd recover the retractability just as promised by Liouville's theorem. $\endgroup$
    – user87745
    Sep 3, 2021 at 15:21
  • $\begingroup$ @mithusengupta123 Liouville's theorem does not apply in such a case. It only applies when you have a Hamiltonian where energy is conserved. This sort of Hamiltonian always applies at a fundamental level, but not if you are ignoring microscopic degrees of freedom. $\endgroup$
    – Dale
    Sep 3, 2021 at 15:24
  • $\begingroup$ @DvijD.C. I understand that. I want to know that if we keep track of the oscillator only and no other degrees of freedom, then even with the exact knowledge of the phase space point of the oscillator at present (say, $Q(0),P(0)$), how/why do we fail to predict which point it came from? I mean, after all, ordinary differential equations in time with enough initial conditions should give a unique answer. So if we have the equation of motion of the damped oscillator, by knowing the "present point" in the oscillator phase space, why can't I find the "past point"? $\endgroup$ Sep 3, 2021 at 15:30
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    $\begingroup$ @jim as long as the Hamiltonian has a conserved energy then the theorem applies. The issue with chaotic systems is that arbitrarily small errors in the inputs lead to arbitrarily large errors in the outputs. $\endgroup$
    – Dale
    Sep 3, 2021 at 16:57
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There are surely many aspects to this question. In particular, the inverse problems are often ill-posed problems. For example, when you reverse time in the heat equation, you have an ill-posed problem because it is extraordinarily sensitive to initial conditions.

See for exemple well-posed problem

Problems that are not well-posed in the sense of Hadamard are termed ill-posed. Inverse problems are often ill-posed. For example, the inverse heat equation, deducing a previous distribution of temperature from final data, is not well-posed in that the solution is highly sensitive to changes in the final data.

For the damped harmonic oscillator, the time reversal leads to an unstable equation (dissipative force are not reversible) which is also very sensitive to the initial conditions. For example, if you start from rest (0,0) the solution x(t) = 0 but if you start from $(ε, 0)$ the solution will be exponentially divergent.

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  • $\begingroup$ "if you start from $(\varepsilon,0)$ the solution will be exponentially divergent" Isn't that exactly what we expect? The time-reversed path in phase space is the actual path traversed in the opposite sense i.e. spiraling outwards from the origin. So the motion is reversible in the sense that we can trace which point the present came from. $\endgroup$ Sep 4, 2021 at 5:48
  • $\begingroup$ Yes, but very small deviations from the initial conditions will lead to huge differences in the prediction of the past of our oscillator. It is the butterfly effect upside down. $\endgroup$ Sep 4, 2021 at 6:41
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The question has to do with how irreversibility emerges from the reversible fundamental laws. The laws of motion are reversible in both classical and quantum mechanics, so the question is not specific to the classical case - even though one could argue that in quantum mechanics the irreversibility is introduced via the measurement.

The origins of irreversible behavior lie in thermodynamics/statistical physics, and express themselves as the law of the increase of entropy, which is actually the manifestation of our lack of knowledge about the details of the system. Had we been able to reverse the velocity of all the atoms, the system would evolve back to its initial state (Loschmidt's paradox)If this never happens, it is because such a level of knowledge and control is impossible.

See this question for more discussions.

Remark: Entropy can be defined in different ways and correspondingly mean different things. Here I mean the increase of the uncertainty of the microscopic state that we are in, as expressed by the Boltzmann formula: $$S=k\log\Omega$$

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  • $\begingroup$ When you say that laws of classical mechanics are reversible, do you also mean that the equation of motion of the classical damped oscillator (which is an effective equation of motion) is also reversible? If not, why? By knowing "the present" of the damped oscillator, I can not uniquely reconstruct "the past" of the oscillator. Isn't it? Also, see my comment at Dvij D. C below the answer by Dale. This is a confusion. $\endgroup$ Sep 4, 2021 at 5:58
  • $\begingroup$ @mithusengupta123 the laws of classical mechanics are conservative - they contain no damping. To correctly introduce damping of the oscillator, one needs to couple it to a bath/reservoir, to which it will lose its energy - this is z thermodynamic process. $\endgroup$
    – Roger V.
    Sep 4, 2021 at 6:03
  • $\begingroup$ I kind of understand that. But I am interested in the effective description of the oscillator where we incorporate damping by hand. Then the question I am asking, whether that effective equation of motion is time-reversible. You may want to see this physics.stackexchange.com/q/664087/164488 $\endgroup$ Sep 4, 2021 at 6:15
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No, because of Chaos Theory.

No, there are systems that are impossible to predict in detail, because even small changes in initial state result in large changes in outcome. These chaotic systems can be as simple as a pair of rod pendulums connected to each other.

double rod pendulum from Wikipedia

Image from Wikipedia

Of course, this is before you even begin to take into account non-deterministic behavior of quantum mechanics, Heisenberg's Uncertainty Principle, or the possibility of a cosmic ray causing a bit to flip in a computer you're modelling the behavior of...

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