Irreversibility Aristoteles law of motion

Question

I am watching the second lecture from the theoretical minimum and do not understand the argumentation of Susskind.

For those who do not want to watch the video. It is about the law of motion from Aristoteles defined as $$\vec{F} = m \cdot \vec{v}$$ whereas $\vec{F}$ determines force, $m$ the mass and $\vec{v}$ the velocity. We can expand the formula to $$F(t) = \dfrac{m(x(t+\Delta)-x(t))}{\Delta}$$ which can be rewritten as $$x(t+\Delta) = \dfrac{\Delta}{m} F(t) + x(t).$$

He creates a theoretical experiment where there is only one axis where there is a spring at point $x(0)$ which pulls every point $x(t)$ towards itself. This means we can define $$F = -kx$$ where $k$ is a constant factor.

We can now put both equations together and get $$x(t+\Delta)=-\dfrac{\Delta}{m}kx(t)+x(t) = x\left[1-\dfrac{\Delta}{m}k\right]$$ which again can be rewritten as $$\dfrac{dx}{dt} = -\dfrac{k}{m}x.$$

We can solve this as differential equation with $$x(t) = x_0e^{-\frac{k}{m}t}.$$

Until here I understand the concept but then he explains why this is not irreversible (you can't predict the past) and I do not get his reasoning. Can someone explain it:

1. Mathematically

2. As an experiment

Answer 1

Suppose that $x=x(t)$ is a solution of the Aristoteles equation of motion. $$m\frac{dx}{dt} = F(t, x(t))$$ Differently to what happens with the Newtonian equation, the other curve $x_1(t):=x(-t)$ is not (in general) a solution of the same equation even if $t$ does not explicitly appear in the expression of the known function $F(t,x)$, i.e. when this function is simply $F(x)$. It is because of the presence of the first derivative. The curve $x_1$ is the temporally reversed evolution of the system with respect to the evolution described by $x(t)$. In this sense Aristoteles' law of motion is not reversible.

Answer 2

All initial conditions tend to converge to the stable equilibrium point $x=0$. Therefore, as you approach the equilibrium point it becomes more and more difficult to distinguish between different solutions (i.e. solutions that started with different initial conditions). An experiment: prepare a glass of water and stir it with a spoon, then let the flow settle under friction. Do it again, stirring in a different way. The final states are the same (nothing moves), so how can you infer the initial condition?

Funny note: Aristotele's mechanics is correct in some physical limits, namely for small creatures (e.g. bacteria) for which friction is much more important than inertia, see Life at low Reynolds number.

In short: Aristotele's mechanics is dissipative, or even better "overdamped"! Dissipative systems are irreversible. This is the "philosophy", see V. Moretti's answer for the math (or this answer).

Answer 3

First, the equation is irreversible in the sense that changing $t\rightarrow -t $ leads to different behavior (unlike Newtonian mechanics). On the other hand, mathematically the past is not unpredictable, as formally you never reach the origin in any finite duration of time. So a math pedant would say Susskind is not right here.

However in practice of course this doesn't matter as very quickly you get to the length-scales much shorter than anyone in a sane mind would expect any classical equations of motion to be applicable.

As a mater of fact Aristotle never proposed any equations of motion of the sort written in the question. This was an after-thought of his followers and critics. He only talked about "natural places" of bodies where they always try to get to. In this sense, a body that found itself in its natural place happily forgets how it got there, which makes Susskind right.

Answer 4

Technically, the rule you described does let you predict the past. Given $\frac{k}{m}$ and $x(t_0)$ at any time $t_0,$ you can find $x(t),$ which gives the position at all future and past times.

There are a few things this law does not do, though.

It doesn't obey Liouville's theorem. The lectures will cover this in more detail, but the principle in classical mechanics is not just that trajectories don't run into each other, but that they don't get "more bunched up". I won't try to explain that in detail here, but roughly speaking, if you have two of these "Aristotelian boxes" that start off with different initial conditions, they continue to be different at all future times. However, they both get extremely close to each other. In principle, if you knew the positions of the boxes with extreme precision, you'd be able to tell the difference between them even at times very large compared to $\frac{m}{k},$ but if you have any finite error, it won't take very long before the boxes become indistinguishable. If they're indistinguishable, there's no going backwards in time; you won't know which box at the present moment is the one that was at large $x$ values in the past.

(In Newtonian physics, there's more to it than just position. Things can get closer and closer in position over time, but if they do, they'll wind up getting further and further apart in momentum in a way that compensates. That complication doesn't apply to the Aristotelian model because position at a given time is the only parameter needed to specify the entire trajectory of the box. With this factor, though, we can say that two Newtonian boxes released and let forward in time will stay the same "distance" from each other as time goes on, if we account for momentum differences as well as position differences.)

Second, if you take a movie of a box undergoing this Aristotelian motion and play the movie in reverse, the box is not undergoing Aristotelian motion any more.

You've stated that the Aristotelian model is $$ \frac{\mathrm{d}x(t)}{\mathrm{d}t} = -\frac{k}{m} x(t).$$

To imagine running this backwards in time, we define $u(t) = x(\tau),$ where $\tau = t_0 - t$ for some $t_0.$ In other words, $u$ is the same motion, but run in reverse. Then $$\frac{\mathrm{d}u(t)}{\mathrm{d}t} = \frac{\mathrm{d}x(\tau)}{\mathrm{d}\tau}\frac{\mathrm{d}\tau}{\mathrm{d}t} = -\frac{k}{m} x(\tau) \cdot (-1)$$

Making one last substitution of $u(t)$ for $x(\tau),$ this is $$\frac{\mathrm{d}u(t)}{\mathrm{d}t} = \frac{k}{m} u(t).$$

So the law for the time reversed motion has a minus sign in it compared to the law for forward motion. Although both are "reversible" laws in the sense that the present predicts both the future and the past, they are different laws. In one case, the spring sucks everything in toward it. In the reverse case, it pushes everything exponentially away.

Contrast this to the Newtonian law $$\frac{\mathrm{d}^2x(t)}{\mathrm{d}t^2} = - \frac{k}{m} x(t).$$

If you make the same substitutions for $u$ and work through the chain rule in the same way, you'll take two time derivatives and get two minus signs, which cancel each other out. You're left with the rule $$\frac{\mathrm{d}^2 u(t)}{\mathrm{d}t^2} = -\frac{k}{m} u(t).$$

This is the same rule as for the forward-time motion. If you take a video of a mass on a spring oscillating back and forth and play that backward in time, it looks exactly the same, except perhaps for a different initial condition.

So the Aristotelian law has one things we could think of as "reversibility" - being able to go into the past, at least theoretically, but it doesn't have other important things we also think of as "reversibility".