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(In Strogatz' Nonlinear dynamics and chaos, page 163)

I've read that any mechanical system of the form: $mx'' = F(x)$ is symmetric under time reversal.

The author notes that if we make the change t -> -t that the second derivative is unchanged while the first derivative is reversed.

My first question is that I see that $x(-t)'$ = $-x(-t)'$ and $x(-t)''$ = $x(-t)''$, however, I'm not quite clear why exactly the mapping t -> -t is of interest to us.

If the idea is to reverse time, then we would want to start at the end of our motion and work backwards along the trajectory we originally came in on. Suppose our original trajectory goes from time t = $t_1$ to t = $t_9$.

So x(-t) will map the largest t value in our original trajectory to the "farthest left" t value on the negative x-axis. Does that mean after we flip our trajectory over the y-axis that we are now interested in the dynamics of a path in the time range t = -9 to t = -1? It seems a bit odd that we would now be looking at a time range over negative time values. Is this a correct interpretation? Is there a better way to look at it?

Thanks.

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  • $\begingroup$ The second derivative is notated via two single quote symbols '' (which displays as $x''$), not a double quote symbol " (which displays as $x"$). $\endgroup$
    – user14972
    Nov 1, 2017 at 4:03
  • $\begingroup$ ? Perhaps related / interesting : Upwind differencing scheme in Finite Volume Method (FVM) $\endgroup$ Nov 9, 2017 at 20:23

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If your concern is the idea of negative time values, you can think of it this way: $t_0$ is some known time point, and $x(t_0)$ is the state at that time.

Finding $x(t)$ for $t > t_0$ means asking: "What will the state be in the future?" Deterministic dynamical systems are such that if we have enough information about the initial condition (in your case, two data points such as $x(t_0)$ and $x'(t_0)$), we know what will be every future state (at least on the maximal interval of existence for the solution).

Finding $x(t)$ for $t < t_0$ means asking: "Given the known state, can we say what the state was in the past?" The answer is yes for any time-reversible system, which includes unique solutions of differential equations such as yours. Think of $t$ as relative $t_0$, whatever that is chosen to be. In general you can assume that $t_0$ is $0$, so that $t$ is just an offset from the known time; thus negative $t$ just means "before the known state" and positive means "after".

As for the symmetry aspect of your question, think of a frictionless pendulum, whose DE is of the form $mx''=F(x)$. For a given angle and a magnitude of the velocity, you could have it on the 'upswing' or on the 'downswing'. In either case, the path it takes will look the same over time, namely: back-and-forth oscillations. Watching a recording of the pendulum, it will be impossible to say whether the video is playing in forward or reverse time, even if you know that there is some known time $t_0$ with a given angle and velocity.

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This "time reversal principle" means the following: If the function $t\mapsto x(t)$ is the solution to the IVP $$mx''=F(x),\quad x(0)=x_0,\quad x'(0)=v_0$$then the function $$t\mapsto \breve x(t):=x(-t)$$ is the solution to the IVP $$mx''=F(x),\quad x(0)=x_0,\quad x'(0)=-v_0\ .$$ This is easily verified using the chain rule. Given the uniqueness of solutions to such IVPs this means that replacing $v_0$ by $-v_0$ in such an IVP is equivalent to running time backwards with the same initial point $x_0$.

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Instead of going from - say - $t=t_0$ to $t=t_1$ you go from $t=t_1$ to $t=t_0$ using variable $t\to T=t_1-t$. The “final condition” of your motion in one case become the initial condition for your motion in the reverse direction.

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  • $\begingroup$ I'm not sure I follow. T1 would go to -T1, which is our initial condition you say now? Aren't we still looking at negative time as I mentioned in the OP? $\endgroup$
    – H_1317
    Nov 1, 2017 at 13:30
  • $\begingroup$ @H_1317 as far as I understand time reversal it means running the solution backwards in time, not for negative $t$'s. $\endgroup$
    – user160660
    Nov 1, 2017 at 15:07
  • $\begingroup$ Ok so then if for t =1 the mapping would dictate we look at what is happening at f(-1), how is what's happening at t= -1 indicative of following the initial path backwards? T=-1 is not guaranteed at all to be doing anything related to the initial trajectory moving forward. $\endgroup$
    – H_1317
    Nov 1, 2017 at 17:45

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