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Briefly, which physical theories are expected to be time reversal invariant? That is, the mapping of $t\to -t$ will not alter the physics.

Even in Classical Mechanics (CM) it is not obvious if time reversal ought to leave things invariant:

  • On the one hand the three laws of Newton seem to be time-agnostic. (e.g. in the second law, reversing the time leaves the accelaration unchanged).
  • On the other hand we know in reality all systems are lossy, e.g. friction is unavoidable, this alone suffices to break any chance of time reversal invariance. This can be easily exemplified, if we take for example a damped oscillator, the Newton's equation of motion is no longer time symmetric, that is: $$m \ddot{x}\propto -\dot{x}-kx \neq m \ddot{x}\propto \dot{x}-kx$$

I guess the last latter point boils down to the second law of thermodynamics. In Quantum Mechanics (QM), from the Schrödinger's equation one can show that a simple mapping of $t \to -t $ will not render the physics invariant, since an additional antilinear mapping of the wavefunction is required, in the sense that only if $\Psi(t)\to \Psi^*(-t)$ is chosen, the equation remains the same.


Questions:

  1. Am I going completely bonkers with any of the above observations?
  2. More importantly, what would it mean for a physical theory to be time reversal invariant? (I don't mean this in the trivial sense, but rather in a grander scheme of things, if we have physical theories that are symmetric in time and those that aren't, how do they stand together in a unified description of nature?) - feel free to discuss the complement of this question if you see it fit.
  3. Do we not expect all physical theories to be continuously symmetric in time due to the conservation of energy?
  4. Is the notion of time in our current understanding of General Relativity and Relativistic Quantum theories in conflict with one another?
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    $\begingroup$ I tried to answer your question, halfway through I realised that nearly every question is too broad or too philosophical in nature. I can say that in physics the pragmatic (and only) view of time is that is that which is measured by a clock. lots of laws are time invariant. So a film of a bouncing ball if done right, will not provide a direction in time, it looks the same if you run it backwards. Your last question, I haven't the expertise, but at the micro level there is a conflict that involves spacetime. When I finish my psychiatric degree, I can then answer your first question. $\endgroup$ – user108787 Sep 26 '16 at 22:50
  • $\begingroup$ Friction dissipates energy. When we say classical mechanics has time reversal symmetry, we do not include the (clearly non-time reversed) dissipative processes. $\endgroup$ – Jon Custer Sep 26 '16 at 22:58
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First of all, it's worth explaining what a time-reversal actually is: given a PDE (such as the Schrödinger equation or Newton's equations for a classical system) which describes the dynamics of the physical quantity $q(t)$. Here, $q(t)$ is the trajectory associated to the initial condition $q(0) = q_0$ and may stand for position $\mathbf{x}(t)$ in classical mechanics or the wave function $\psi(t)$. (For simplicity, I have arbitrarily chosen the initial time to be $0$.)

A time-reversal symmetry is a transformation $T$ acting on solutions so that \begin{align*} (T q)(-t) = T \bigl ( q(t) \bigr ) \end{align*} holds for all initial conditions; it is not the map $t \mapsto -t$. The above equation means the following: on the left side you evolve the trajectory associated to the initial condition $T q_0$ backwards in time. On the right, you apply $T$ to the trajectory $q(t)$ associated to initial condition $q_0$.

In non-relativistic quantum mechanics for a particle without spin, time-reversal symmetry is given by complex conjugation, i. e. $T \psi = \overline{\psi}$. If $T$ commutes with the Hamiltonian $H$ (e. g. in the absence of magnetic fields), then the Schrödinger equation has time-reversal symmetry: $T \, H = H \, T$ implies \begin{align} T \, \mathrm{e}^{- \mathrm{i} t H} = \mathrm{e}^{- \mathrm{i} (-t) H} \, T \end{align} so that for a solution $\psi(t) = \mathrm{e}^{- \mathrm{i} t H} \psi_0$ of the Schrödinger equation to initial condition $\psi_0$, we have \begin{align*} T \bigl ( \psi(t) \bigr ) &= \overline{\psi(t)} = T \, \mathrm{e}^{- \mathrm{i} t H} \psi_0 = \mathrm{e}^{- \mathrm{i} (-t) H} \, T \psi_0 = (T \psi)(-t) . \end{align*} Note that it is not just losses which may break time-reversal symmetry of physical equations, magnetic fields are another (which evidently conserve energy). There are also cases where you have more than one operation which implements time-reversal on a physical level. If, for instance, your quantum Hamiltonian has a chiral-type symmetry, i. e. if there exists a linear $T$ which anticommutes with your Hamiltonian ($T \, H = - H \, T$), then $T \, \mathrm{e}^{- \mathrm{i} t H} = \mathrm{e}^{- \mathrm{i} (-t) H} \, T$ still holds (this time the sign flip is not due to the antilinearity, but the anticommutativity).

Another such instance are Maxwell's equations where the transformation $T : (\mathbf{E} , \mathbf{B}) \mapsto (\mathbf{E} , -\mathbf{B})$ and $T : (\mathbf{E} , \mathbf{B}) \mapsto \bigl ( \overline{\mathbf{E}} , -\overline{\mathbf{B}} \bigr )$ both reverse the arrow of time. (Adding complex conjugation does make sense if one works with complex electromagnetic fields.)

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  • $\begingroup$ Thanks a lot for your answer, very helpful. I understand now that I should think of TR as a symmetry transformation just like that of rotation when a system is rotationally invariant. What I'm still struggling with is why the transformation T is not unique. Finally, would you be so kind to tackle question 2 just briefly, namely the bigger picture perspective. Many thx again. $\endgroup$ – user929304 Oct 6 '16 at 16:56
  • $\begingroup$ I just noticed that I had forgotten to garnish $\mathbf{B}$ with a - sign, sorry about that. If you have two symmetries which reverse the arrow of time, keep in mind that they don't map onto the same state. You could multiply any time-reversal symmetry (TRS) by -1, and obtain “another” TRS. What symmetry officially gets the title depends also on convention. E. g. in quantum mechanics, TRSs are by definition anti-unitary, commuting operators, so linear, anticommuting (pseudo-)symmetries are called chiral symmetries. For classical equations, this terminology may not make sense, though. $\endgroup$ – Max Lein Oct 7 '16 at 12:15
  • $\begingroup$ Nvm previous comments, I somehow ignored antilinearity. $\endgroup$ – udrv Apr 14 '17 at 12:42
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As for your first question, yes a bit bonkers. The best way to think of it without getting so complicated is to understand that time symmetry is a micro-symmetry, valid for all of the basics forces of nature, except the weak forces. just like CP.

For the macroscopic physics entropy and losses due to heat and friction enter in, and entropy principally provides an arrow of time as to what is the past and the future.

Worthwhile nothing that entropy, like anything from statistical mechanics, depends on the level of detail of one's model. You could try to model the microscopic interactions and kinetic energies of the molecules that are heated by friction, and if you had a small enough macroscopic area, or some way of simplifying things, you could try work out all the details, with a very very large computer. Entropy refers to large enough systems of particles and forces or fields that practically you can not do the microscopic calculations for all of them.

I'd suggest take a course, even if basic, on statistical mechanics and you'll pick up the way thermodynamic quantities like entropy and temperature arise from microscopic elementary mechanics and then more complex physics. And it can help you see beyond the words at this level.

2nd question: no differently than other symmetries and lack of them. The weak force violates CP symmetry, the others don't. Same for time symmetry. The effects are relatively small. There is still a hope/expectation that the prevalence of particles over antiparticles in the universe is due to this. The more 'in the grander scheme of things' answer to your question I think needs to await a more specific question. Try posting a specific paradox you can think of by having that mix.

3rd) Well, in general,relativity you can have non time symmetric spacetimes, such as our present cosmological model. Energy is not conserved.

4th)Quantum Field Theory has energy conserved, and time a symmetry in Minkowski spacetime. Time reversals do not affect that. General Relativity, see the 3rd answer. There has not been a way to merge quantum theory and general relativity, i.e., no accepted,theory of quantum gravity.

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  1. No, these are good observations.

  2. In order for a theory to be time-reversal invariant at the micro level, as you say, you need that the fundamental (not phenomenological) differential equations describing its time evolution to only contain even time derivatives. As you point out, if you want the theory to also be TR invariant at the macro level, you need to defang the second law of thermodynamics, which introduces a bias toward increasing entropy. The reason that it does so is because the initial state of the universe at the Big Bang appears to have been an extremely (i.e. highly non-generically) low-entropy state, and so the entropy has been increasing ever since. I.e. the laws of physics are perfectly TR invariant (with the slight and probably unimportant exception of the weak force), but the initial/boundary conditions strongly break the symmetry - the initial state of the universe was extremely low-entropy and the final state will presumably be extremely high-entropy. This non-generic initial condition (which is still not fully understood, although cosmic inflation may help explain it) explains why we observe dissipative processes today! (In fact in explains why we observe any structured phenomena today, instead of just a maximally entropic mess.)

So in order for your theory to be TR invariant at both the micro and macro level, you need both the fundamental differential equations to be TR invariant, and also you need the system to be in a maximum-entropy state. (Or you could have a highly non-generic system that doesn't thermalize, e.g. a completely closed integrable system like a quantum Newton's cradle.)

  1. The Lagrangian describing the three non-gravitational forces is time-independent, so energy (more generally, the stress-energy tensor) is indeed (classically) conserved for these processes. But in general relativity, whether or not we can assign a conserved energy-like quantity depends on whether or not the metric possesses a timelike Killing field. The FRW metric that approximately describes our universe does not have such a field, so energy is not conserved. Roughly speaking, dark energy "appears out of nowhere" as the universe expands.

  2. To some extent, yes. For example, the unitarity of time-evolution in quantum mechanics implies that all processes should be (microscopically) reversible by crossing symmetry, while in classical GR geodesics can terminate at singularities or fall behind event horizons and their "information gets lost," making time-reversibility a tricky issue that is still not understood today. But this question is too broad to address completely.

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  • $\begingroup$ Anyone know how to stop the site from automatically formatting my answer as a list? It messes up the formatting of multi-paragraph entries. $\endgroup$ – tparker Oct 3 '16 at 15:49
  • $\begingroup$ Thanks a lot, very interesting discussions. What would be the repercussions from a physical point of view of the macroscopic laws of physics were also to be TR invariant? Finally, if time allows, how does the time reversal picture shift when going from CM to GR, where t is no longer a parameter but a dimension on its own? Many thanks for any inputs. $\endgroup$ – user929304 Oct 6 '16 at 17:05
  • $\begingroup$ (1) If the macroscopic laws of physics were TR invariant, then the second law of thermodynamics would not apply, and indeed the notion of time itself would stop being that useful a concept because everything would necessarily be in a macro equilibrium. (Very roughly, time would become kind of like another spatial dimension, in the sense that motion forward and backward in time would be qualitatively identical, just as motion left and right are.) Presumably the universe would be a maximal entropic mess with no organized structure whatsoever - e.g. a gas of photons. $\endgroup$ – tparker Oct 7 '16 at 2:17
  • $\begingroup$ (2) The Einstein equations are symmetric under time reversal. However, there is certainly an empirical asymmetry, which is that gravity (excluding dark energy) tends to pull everything together rather than push them apart. This can be formalized by placing various restrictions on the stress-tensor - basically, requiring that all masses be positive rather than negative so that things attract rather repel. $\endgroup$ – tparker Oct 7 '16 at 2:23
  • $\begingroup$ Even with these conditions imposed, the micro laws remain TR invariant, but if there's any kind of energy dissipation, then these energy conditions will result in a macro time asymmetry where things tend to "clump together." $\endgroup$ – tparker Oct 7 '16 at 2:24

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