Time Reversal Symmetry in Classical Physics

[hokhani]

TL;DR

I don't understand how does time-reversal applies to the free-fall problem.

I try to justify time-reversal symmetry in a very simple classical problem; Free Fall. The position, $x$, and the velocity, $v$ are obtained versus time from the equation $-g=\ddot x$. So, if we consider the primary conditions as $t_0,x_0,v_0$ it is clear that, $x(t)=\frac{1}{2}(-g)(t-t_0)^2+v_0(t-t_0)+x_0$ and $v(t)=-g(t-t_0)+v_0$ give the position and velocity at any time. Plesase see the attached figure. AT the point (1) we have$$x(t_1)=x_1=\frac{1}{2}(-g)(t_1-t_0)^2+v_0(t_1-t_0)+x_0$$ and $$v(t_1)=v_1=-g(t_1-t_0)+v_0.$$ We expect by applying time reversal to obtain $x(-t_1)=x(t_1)$ and $v(-t_1)=-v(t_1)$. However, inserting $-t_1$ instead of $t_1$ doesn't neccesarily fulfill this requirement:
$$x(-t_1)=\frac{1}{2}(-g)(-t_1-t_0)^2+v_0(-t_1-t_0)+x_0$$ and $$v(-t_1)=-g(-t_1-t_0)+v_0.$$ I would like to know what is my mistake?

 

[vanhees71]

It's a bit a subtle point you stumbled on. Time reversal doesn't simply mean that you make $t \rightarrow t'=-t$, but you also have to start with the time-reversed initial condition, i.e., $x_0'=x_0$ and $v_0'=-v_0$. Also accelerations don't change under time-reversal, i.e., $g'=g$. Now take your solutions (for convenience I set $t_0=0$)
$$x(t)=-\frac{g}{2} t^2 + v_0 t +x_0.$$
Now let's look at the time-reversed process. Just make the above transformations of both the time variable and the initial conditions. Then you get
$$x'(t')=x(t)=x(-t')=-\frac{g'}{2} t^{\prime 2} -v_0 t' + x_0'=-\frac{g}{2} t^{\prime 2} + v_0' t' + x_0',$$
which is the same as the original solution. This reflects time-reversal invariance of your mechanical system.

 

[hokhani]

It's a bit a subtle point you stumbled on. Time reversal doesn't simply mean that you make $t \rightarrow t'=-t$, but you also have to start with the time-reversed initial condition, i.e., $x_0'=x_0$ and $v_0'=-v_0$. Also accelerations don't change under time-reversal, i.e., $g'=g$. Now take your solutions (for convenience I set $t_0=0$)
$$x(t)=-\frac{g}{2} t^2 + v_0 t +x_0.$$
Now let's look at the time-reversed process. Just make the above transformations of both the time variable and the initial conditions. Then you get
$$x'(t')=x(t)=x(-t')=-\frac{g'}{2} t^{\prime 2} -v_0 t' + x_0'=-\frac{g}{2} t^{\prime 2} + v_0' t' + x_0',$$
which is the same as the original solution. This reflects time-reversal invariance of your mechanical system.

Many thanks for your reply.
However, I still don't know the time evolution of the system in the time-reversal state. Better to say, consider the following figure. The left panel shows the motion in the direct time and the right panel, in the time-reversal states. What is the time evolution of the system in the right panel? The particle must fall from the point with $-t$ to the point $-t_0$ or vice versa?
In other words, as shown in the schematic clock at the rightmost figure, which direction is the time evolution in the time-reversal state (between the two times $-t$ and $-t_0$) ?

 

[vanhees71]

Time evolution is always forward i.e. from smaller to larger values of $t$, i.e., your time-reversed situation describes free fall of a particle starting at time $t$.

 

[hokhani]

Time evolution is always forward i.e. from smaller to larger values of $t$, i.e., your time-reversed situation describes free fall of a particle starting at time $t$.

So, it seems that time-reversal doesn't give a new solution to equation $x\ddot =-g$, and only changes the initial conditions. We could set the initial conditions at the top with a downward velocity, and then exactly the same motion would occur. In other words, Time reversal doesn't present a new motion!

 

[vanhees71]

That's the point! It leads again to a solution of the equations of motion, which means that Newton's law for the given force law is invariant under time reversal.

 

[wrobel]

Time reversal symmetry means exactly the following trivial fact: if $x(t)$ is a solution to the equation $\ddot x=f(x),\quad x\in\mathbb{R}^m$ then $x(\hat t-t)$ is also a solution to the same equation. Sure $x(t)$ must be defined on an interval consistent with the constant $\hat t$ and bla-bla other evident math details

 

[vanhees71]

Well, it's mathematically trivial, but physicswise it's not. First of all, how do you explain that in everyday life there's an obvious breaking of this symmetry. It's not as trivial as it sounds since the "thermodynamic arrow of time" from Boltzmann's H Theorem is not an explanation but an assumption, because you assume the directedness of time in deriving the Boltzmann equation via making the assumption of "molecular chaos". I.e., all you prove is that the "thermodynamic arrow of time" is identical with the assumed "causal arrow of time".

In electromagnetism you also have to make the assumption of directedness of time and then choose the retarded solution over all other possible solutions of Maxwell's equations. The argument is pretty similar to the argument in deriving the H theorem, i.e., this defines a "radiation arrow of time", but it's again only using the assumption of the "causal arrow of time".

The only known violation of time-reversal invariance is due to the weak interaction, implying together with the CPT theorem of local QFTs that CP is violated in the standard model too. The only trouble is, it's not enough to explain our very existence (though there was some hope from the recent neutrino result by T2K):

https://www.nature.com/articles/s41586-020-2177-0
https://www.quantamagazine.org/neutrino-asymmetry-passes-critical-threshold-20200415/

 

[hokhani]

That's the point! It leads again to a solution of the equations of motion, which means that Newton's law for the given force law is invariant under time reversal.

How about the origin of the time, $t=0$. Is it arbitrarily chosen?

 

[wrobel]

So, it seems that time-reversal doesn't give a new solution to equation x¨=−gx\ddot =-g , and only changes the initial conditions

different initial conditions provide different solutions (existence and uniqueness theorem)

 

[vanhees71]

How about the origin of the time, $t=0$. Is it arbitrarily chosen?

Sure, what you count as $t=0$ is just an arbitrary setting of your clock. In Newtonian mechanics time is translation invariant (which leads to energy conservation due to Noether's theorems).

 

[hokhani]

Sure, what you count as $t=0$ is just an arbitrary setting of your clock. In Newtonian mechanics time is translation invariant (which leads to energy conservation due to Noether's theorems).

So, I conclude that:
If we start at the initial time $t=0$, then, follow the time direct (time reversal) motion, we would go toward later times by time evolution. However, the time-reversal motion would take the system behavior towards the last times.
Is this conclusion correct?