# Time Reversal Symmetry in Classical Physics

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• hokhani
In summary: 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...).
hokhani
TL;DR Summary
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 teh 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?

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.

haushofer, hokhani and (deleted member)
vanhees71 said:
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.
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##) ?

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##.

etotheipi
vanhees71 said:
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.

PeroK and hokhani
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

etotheipi and 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/

vanhees71 said:
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?

hokhani said:
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)

hokhani said:
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
vanhees71 said:
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?

Last edited:

## 1. What is Time Reversal Symmetry in classical physics?

Time Reversal Symmetry is a fundamental concept in classical physics that states that the laws of physics remain unchanged when time is reversed. This means that if we were to reverse the direction of time in a physical system, the behavior of that system would still follow the same laws and equations.

## 2. How does Time Reversal Symmetry affect the behavior of physical systems?

Time Reversal Symmetry has a significant impact on the behavior of physical systems. It allows us to predict the future behavior of a system by observing its current state, as well as to understand its past behavior by observing its current state. This is because the laws of physics are symmetric in time, meaning they are valid in both the past and the future.

## 3. Are there any exceptions to Time Reversal Symmetry in classical physics?

While Time Reversal Symmetry is a fundamental concept in classical physics, there are some exceptions to this rule. One example is the phenomenon of irreversible processes, such as the flow of heat from a hot object to a cold object. These processes violate Time Reversal Symmetry because they cannot be reversed in time.

## 4. How is Time Reversal Symmetry related to the concept of entropy?

Entropy is a measure of the disorder or randomness in a system. Time Reversal Symmetry is closely related to entropy because it states that the overall entropy of a closed system will remain the same over time. This means that the disorder of a system will not decrease or increase when time is reversed, but will remain constant.

## 5. Can Time Reversal Symmetry be applied to all physical systems?

While Time Reversal Symmetry is a fundamental concept in classical physics, it cannot be applied to all physical systems. In particular, it does not hold true in systems that involve quantum mechanics, as the laws of quantum mechanics are not symmetric in time. Additionally, systems that involve strong gravitational fields, such as black holes, may also violate Time Reversal Symmetry.

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