Time reversal (as it's normally understood) transforms this motion to another possible motion, that of a ball rolling up an inclined plane until coming to rest.
In other words, this picture of time reversal reverses the time-ordering, preserves position, and reverses momentum.
The philosophical question for us is: what does time reversal generally mean? In particular, we'd like to know which transformations count as time reversal, as opposed to any other transformations. Well, here's one interesting way to produce an answer. Suppose the following principle is true of the world.
Free Motion Symmetry. In the absence of forces of interactions, when a system contains only free particles or fields, the laws of nature are time reversal invariant.Nature need not be time reversal invariant in general. But adopting this principle means that when we turn off all interactions, there is no question -- if a free motion is allowed by nature, then so is its time reverse. I think there is good reason to believe this principle. But whether there is or not, it's interesting to observe that it provides a unified way to recover the standard time reversal operators. For example:
Theorem 1. Suppose that Hamilton's equations are time-reversal invariant for the free-particle Hamiltonian, and that T is a linear involution (T2=1). Then T is characterized by one of the following:In other words, free motion symmetry (together with the assumption that T is an involution or reversal) is enough to pick out the two time-reversal operators of classical mechanics. The first is the standard time reversal operator. The second is the space-and-time reversal operator.
Tq = q and Tp = -p
or
Tq = -q and Tp = p
where q and p are the canonical position and momentum variables, respectively.
As it turns out, this same method recovers the standard (but more unusual looking) time-reversal operator in quantum mechanics as well. In particular, we have the following.
Theorem 2. Suppose that T is a Hilbert space symmetry that commutes with the free-particle Hamiltonian. If the Schrödinger equation with this Hamiltonian is time-reversal invariant, then T is antiunitary.In other words, assuming free motion symmetry in a system characterized by a time-independent Hamiltonian is enough to guarantee that T is antiunitary -- which is the standard characterization of the quantum time reversal operator.
For more on this approach (and the very elementary proofs of the theorems), see my recent draft: How to time-reverse a quantum system, or my post on how to visualize why time-reversal involves conjugation in quantum mechanics.
13 comments:
>> when a system contains only free particles or fields
One problem with this is that non-interacting
particles and fields cannot be observed.
Strictly speaking you cannot even know if "a system contains only free particles or fields".
That may be true... but why is this a problem? It seems we can still have good reason to believe the Free Motion Symmetry principle.
But what would be those 'good reasons'?
Obviously, we have no experience from/with free particles directly (we can only extrapolate from (weakly) interacting particles).
So I would think you either formulate the principle with interactions and then consider the limit of turning them off (by why would you do that if you already have the principle for the more general case).
Or you have to live with the fact that you are making statements about something which cannot be observed in principle.
(by why would you
should have been
(but why would you
By the way, in order to show that this is not idle sophistry, consider what you mean with 'free particle' in the context of general relativity.
PS: sorry for posting 3 comments to express one thought...
Wolfgang: no worries! Here's a bit of a long response.
Suppose I grant your point, that we could never empirically verify Free Motion Symmetry. We can still have reason to believe it, in the same way that we believe the spacetime manifold is smoothly differentiable -- another empirically unverifiable claim. We accept both principles, because they allow us to construct useful theories and make useful predictions that would otherwise be impossible.
One reason we need Free Motion Symmetry is for the study of whether or not interactions have (or don't have) some symmetry like parity or time reversibility. Without FMS, a parity asymmetry in (say) a W-boson interaction could always be attributed to an asymmetry of the underlying kinematics, and not the interaction itself.
Another reason we need FMS if to apply Wigner's method of deriving quantum numbers. If Wigner hadn't assumed that the amplitudes described by a free quantum system remain fixed under spacetime symmetries, his whole procedure would come undone. But he did assume it, and this procedure (it seems) led to a very fruitful program in the foundations of quantum theory.
Does the principle make sense in GR? I haven't worked this out carefully yet, but I suspect it does. Just take a 'free particle' to be a test particle, and its equations of motion to be the geodesic equations. Then free motion symmetry is a theorem -- it just says that the geodesic equations are covariant under the symmetries (isometries) of the background spacetime.
... Or is my sophistry idle? ;)
>> Just take a 'free particle' to be a test particle
i think you would have to be very careful about what you assume about the test particle (and its mass and the bakground geometry).
True, and point taken. But the test particle idealization is just a convenience in general relativity. I could always choose a particular world line in a real evolving matter field -- Geroch showed that such a worldline will be a geodesic, given some reasonable assumptions about the matter field.
thinking about the q.m. case, i think there is actually a counterexample here:
first we 'turn off' gravity by setting G=0, which will decouple spacetime geometry from matter. The field equation for the geometry are now Rij = 0 , in other words the geometry is Ricci flat. One solution is the Schwarzschild geometry.
As for the matter we consider a non-interacting quantum field, so we have your 'free particles' case.
But this means you have the situation originally considered by Hawking (non-interacting quantum field on a Schwarzschild geometry). So you will find Hawking radiation and run into the information loss problem. This means that unitarity breaks and your time reversal operator must be problematic.
so this illustrates my original point that the idea to 'turn off' all interactions is somewhat ambiguous (and we cannot have direct experience with free particles).
perhaps it can be done in a fully consistent quantum theory of everything (string theory?), but to repeat a previous point, why would one want to do this if we already have the consistent theory with interactions ...
Nice example! I'm still not convinced this is a problem for free motion symmetry though. The natural vacuum state around a Schwarzschild black hole undergoes an irreversible thermal process. But that by itself doesn't violate free motion symmetry. Do we really have a free field in the presence of a black hole? There seem to be two situations, neither of which is a problem for FMS.
(1) Outside event horizon, it's clear that we can have a free quantum field. And in that region, we do expect the equations of motion to have the symmetries of the background spacetime. We even use this fact in the usual derivations of Hawking radiation, when we assume that the equations of motion are "Bogolyubov transformation invariant," whatever they are. Here irreversible processes are not a problem.
(2) Inside the event horizon, it's not so clear that we have a free field. Indeed, one way to put the problem of quantum information loss is that the quantum field seems to interact with the spacetime singularity (or its analogue in quantum gravity) in a way that we don't yet understand. The inverse Bogolyubov transformation doesn't exist here, and time-reversibility may break down -- but should be prepared for this kind of symmetry breaking to occur in the presence of interactions.
Can you generalize your Theorem 2 if there are interactions?
I looking in your draft, I don't see a rigorous proof of theorem 2. (No rigorous proof for your proposition 1?)
This isn't a complaint, just that if you wrote out the rigorous proof in detail, maybe you will find a way to add various interactions to it where the argument still holds. (Only a guess.)
Good luck with the time issues.
Joseph,
The proofs are in the appendix of the paper. I'm making this argument in the context of ordinary quantum mechanics, because it's simplest to see how it works there -- but it generalizes naturally to quantum field theory.
Adding interactions isn't a problem -- the antiunitarity of T shouldn't depend on one's Hamiltonian. T is a general transformation of the motions allowed by a theory. So, establishing that T has a certain property for the free-particle case is enough to establish it has the same property in interacting cases as well.