Does it make sense to say that the action is even or odd under time reversal?

The action of a system in mechanics is an integral over time defined as $$S[x(t)]=\int\limits_{t_1}^{t_2}L(x,\dot{x},t)dt.$$ Here, the time $$t$$ is integrated making the left hand side depend only on the $$x$$-values at the endpoints of the integral i.e. $$x(t_1)$$ and $$x(t_2)$$. I was thinking if it is correct to ask about the time-reversal property of the action? I mean is it meaningful to say that the action is even or odd under time-reversal? If not, what would be time-reversal symmetry in classical mechanics?

EDIT In response to G. Smith's comment, I want to point out that the action of a free particle $$S$$ can be computed to be $$S=\frac{1}{2}mv_0^2(t_2-t_1)$$ where $$v_0$$ is the velocity of the particle for all times. So it is not really a function of $$t$$.

• Are you allowing for external time dependence? – Qmechanic Jun 16 at 15:40
• If you mean explicit time-dependence, yes. Actually, both. – mithusengupta123 Jun 16 at 15:52
• The action can be even or odd under time reversal but it doesn’t have to be either one, any more than a function has to be even or odd. – G. Smith Jun 16 at 19:39
• @G.Smith That's true. But since here t is integrated, I don't know whether it makes sense to say that S is even, odd or neither under time reversal. – mithusengupta123 Jun 21 at 21:29
• “The time $t$ is integrated making the left hand side depend only on the $x$-values at the endpoints of the integral i.e. $x(t_1)$ and $x(t_2)$.” This is not true, unless $L$ is a total time derivative, which is usually not the case. You are missing the point that the action depends depends on the entire trajectory between $t_1$ and $t_2$. – G. Smith Jun 21 at 21:41

This is a good question. I had thought about this in terms of parity but I believe the same logic will apply. The action (and the lagrangian) determine the equations of motion and the equations of motion are either time reversal symmetric or time reversal asymmetric. We can look at this through the Euler Lagrange equation, $$\frac{d}{dt}\frac{\partial \mathcal{L}}{\partial \dot q}=\frac{\partial \mathcal{L}}{\partial q}.$$ Now if the Lagrangian is even under time reversal (I will denote the time reversal operations as $$\mathcal{T}$$, \begin{align} \mathcal{T}\frac{d}{dt}\frac{\partial \mathcal{L}}{\partial \dot q} &= \mathcal{T}\frac{\partial \mathcal{L}}{\partial q}\\ (-1)\frac{d}{dt}(-1)\frac{\partial}{\partial \dot q} \mathcal{T}\mathcal{L}&= \frac{\partial}{\partial q}\mathcal{T}\mathcal{L}\\ \frac{d}{dt}\frac{\partial \mathcal{L}}{\partial \dot q}&=\frac{\partial \mathcal{L}}{\partial q}. \end{align} However if the lagragian is odd under time reversal we get, \begin{align} (-1)\frac{d}{dt}(-1)\frac{\partial}{\partial \dot q} \mathcal{T}\mathcal{L}&= \frac{\partial}{\partial q}\mathcal{T}\mathcal{L}\\ -\frac{d}{dt}\frac{\partial \mathcal{L}}{\partial \dot q}&=-\frac{\partial \mathcal{L}}{\partial q}\\ \frac{d}{dt}\frac{\partial \mathcal{L}}{\partial \dot q}&=\frac{\partial \mathcal{L}}{\partial q}. \end{align} We still get time reversal symmetric motion! This has been asuuming at all of the terms in the lagrangian flip the same way under time reversal. However if this is not the case we can get time reversal asymmetric motion. We can look at a simple example where our lagrangian has 2 parts, $$\mathcal{L}(q,\dot{q}) = A(q,\dot{q}) + B(q,\dot{q})$$ Where $$A$$ is even and $$B$$ is odd. Now our equations of motion will look like, $$\frac{d}{dt}\frac{\partial}{\partial \dot{q}}\left(A+B \right) = \frac{\partial}{\partial q}\left(A+B\right).$$ Now the motion is not time reversal symmetric as under time reversal we get, $$\frac{d}{dt}\frac{\partial}{\partial \dot{q}}\left(A-B \right) = \frac{\partial}{\partial q}\left(A-B\right).$$ This looks a fair amount like weak interactions in the standard model (in this case breaking parity not time reversal). The terms $$\bar{\psi}\gamma^\mu\psi$$ and $$\bar{\psi}\gamma^\mu\gamma^5\psi$$ are odd and even respectively under parity transformation. Weak interactions involve both terms in the form $$\bar{\psi}\gamma^\mu(1-\gamma^5)\psi$$, and it is in the interference of these two terms that we get parity violation.

How this applies to the action itself is that the action is either symmetric/antisymmetric or asymmetric under the transformation. If it is asymmetric then the motion will not respect the transformation. You noted in the question an action that could explicitly depend on time, and I have not addressed that. If I remember correctly the Euler Lagrange equations have to be modified and you could then rework the arguments I presented here.

• This is clearly not a rigorous proof but more of a heuristic argument/example. Hopefully it at least shows some insight into the problem though. – TEH Jun 16 at 17:55
• "the action is either symmetric/antisymmetric or asymmetric under the transformation. " How would you prove that? In the action, time is integrated. – mithusengupta123 Jun 16 at 22:03
• Not a proof and more of an argument, but time reversal acting on the action is an operation that maps $\mathbb{R} \rightarrow{} \mathbb{R}$, this gives $\mathcal{T}S = S'$. Now either $S = S'$ which we call symmetric or $S \neq S'$ which we call asymmetric. I'm not sure what other outcome is possible. – TEH Jun 16 at 22:29
• antisymmetric or asymmetric? – MannyC Jun 28 at 3:40

Yes: it makes perfect sense to say that the action is even/odd under time reversal.

A canonical example of an action that is even under time-reversal is the Yang-Mills action. A canonical example of an action that is odd under time-reversal is the Chern-Simons action1. A trivial example of an action that is neither odd nor even is the action that contains both a Yang-Mills part and a Chern-Simons part.

Note that these terms exist in any number of odd dimensions and, in particular, in $$0+1$$ dimensions, i.e., point mechanics.

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For a more down-to-earth example, consider the action $$S[q]=\int_{\mathbb R}\dot q^n$$ which is even (resp. odd) under time-reversal if $$n$$ is even (resp. odd). Indeed, the action satisfies $$S[\tau^*q]=(-1)^nS[q]$$, where $$\tau(t)=-t$$.

1: This leads to the well-known Redlich anomaly: a system that contains fermions and no Chern-Simons term is classically parity (equiv. time-reversal) invariant, but it is not as a quantum theory, because the fermions (may) generate a Chern-Simons term.

• Can you give examples from classical mechanics? It would be easier for me. – mithusengupta123 Jun 25 at 19:55
• @mithusengupta123 Everything I said (except for the footnote) is classical. There are analogous quantum statements, but you don't have to worry about them here if you don't want to. – AccidentalFourierTransform Jun 25 at 21:23
• Still your examples are too advanced. Could you please explain how does the action of a particle, free or in a potential, be even, odd or neither under time reversal if $t$ is integrated. Thanks – mithusengupta123 Jun 25 at 23:40
• @mithusengupta123 The fact that $t$ is integrated is irrelevant. Symmetries concern endomorphisms of a function space, not named variables. You could call the integration parameter $\tau$ instead of $t$; that wouldn't change the physics. Have a look at physics.stackexchange.com/a/403359/84967 and make sure you understand what it means for $S$ to be symmetric under some transformation. – AccidentalFourierTransform Jun 25 at 23:49