Explanation of a spring using "Artistotle's law of motion", $\vec{F}=m\vec{v}$ So I was watching Susskind's Classical Mechanics lectures and I didn't understand something in the second lecture.
He was telling about Aristotle's Law of motion which is $$\vec F = m\vec v.$$
He applied this law on a spring-block system and got this function of position.
$$x(t)=x(0)\cdot e^{-\tfrac{k}{m}t}.$$
He then said that this law is not valid under classical mechanics as it does not help us to predict the past because after some time the particle would just be standing at the origin and predicting where it started from is not possible with finitely accurate measurements.
This is what I did not understand. There are many such laws like for example a damped harmonic motion which consist of an exponential decay. Are they also not valid?
 A: The equation $F = m v$ is not time-symetrical, i.e. reversing the sign of $t$ gives a different equation because of the appearence of velocity.  The equation $F = m a$ is second order in $t$, so it doesn't change under time inversion $t \Rightarrow -\, t$, but only if the left member ($F$) doesn't include terms with dependance on velocity (like friction).  If there's a friction, then $F$ contains a non-symetrical term under time-inversion.  Friction implies a loss of information.
If you have a strict exponential damping: $x(t) = x_0 \, e^{- \lambda t}$, then measuring position and velocity at time $t_1$ gives this system of equations:
\begin{align}
x_1 &= x_0 \, e^{- \lambda t_1}, \tag{1} \\[12pt]
v_1 &= -\, \lambda \, x_1. \tag{2}
\end{align}
You then knows $x_1$ and $v_1$ (so you know $\lambda$ from (2)), but you don't know time $t_1$ and want to retrodict the initial position $x_0$ (at time $t_0 = 0$).  Equation (1) gives you only one equation for two unknows.  You need more information to retrodict the past.  Measuring position and velocity again at time $t_2$ gives you two new equations:
\begin{align}
x_2 &= x_0 \, e^{- \lambda t_2}, \tag{3} \\[12pt]
v_2 &= -\, \lambda \, x_2. \tag{4}
\end{align}
Equation (4) is useless.  Combining (1) and (3) gives this, where $\Delta t = t_2 - t_1$ is known:
\begin{equation}\tag{5}
x_2 = x_0 \, e^{- \lambda (t_1 \,+\, \Delta t)} = x_1 \, e^{- \lambda \, \Delta t},
\end{equation}
so there's nothing new here, and you can't find $x_0$!
In other words:  The pure exponential function has no memory!  This fact is very important for the radioactive decay and for statistical theory.
