Thoughts on testing reversibility of Aristotle's law of motion in Susskind's Classical Mechanics Theoretical Minimum
The following is my attempt to decipher what Leonard Susskind communicates in Lecture 3: Dynamics of Theoretical Minimum: What You Need to Know to Start Doing Physics, when he formulates “Aristotle’s Law of Motion”, and shows that this law doesn’t hold when time is reversed.
Susskind defines Aristotle’s Law of Motion as:
$$\vec{F} = m \vec{v}$$
from which he gets:
$$F(t) = m \frac{dx}{dt} \tag{1}$$
He then checks if the law is reversible. He claims we can do it as follows:
Everywhere you see time in the equations, replace it with minus time. That will have the effect of interchanging the future and the past. Changing t to -t also includes changing the sign of small differences in time. In other words, every Δt must be replaced with -Δt. In fact, you can do it right at the level of the differentials dt. Reversing the arrows means changing the differential dt to -dt.
He applies this logic, and arrives at:
$$-F(-t) = m \frac{dx}{dt} \tag{2}$$
This derivation was not obvious for me to follow. This led me to:
Later, he concludes with:
The implication is simple: The reversed equation of motion is exactly like the original, but with a different rule for the force as a function of time.
I got hang up over the word “implication”. I was not able to see:
I spent over 20 hours thinking about this problem. Mostly about question #2. In this article, I am documenting my findings.
Question #1: How do we arrive at equation (2) from equation (1), and what does the right hand side of equation (2) refer to?
To follow Susskind’s logic, I first made the notation more obvious to myself. I introduced subscripts to functions, to know whether I was referring to forward-time, or reversed-time motion:
fwddescribes functions in forward-timerevdescribes functions in reversed-time
Position at time t, with forward-time is given by:
$$x_{fwd}(t)$$
Velocity is, by definition, the rate of change in position with respect to time:
$$v_{fwd} = \frac{d}{dt}x_{fwd}(t) \tag3$$
Aristotle’s law of motion relates object’s velocity, its mass, and the force applied to it:
$$F_{fwd}(t) = m\frac{d}{dt}x_{fwd}(t) = mv_{fwd}(t) \tag4$$
With reversed-time, Aristotle’s law of motion has the following form:
$$F_{rev}(t) = m\frac{d}{dt}x_{rev}(t) = mv_{rev}(t) \tag5$$
Position at time t with reversed-time, is defined by the position at time -t with time going forward:
$$x_{rev}(t) := x_{fwd}(-t)$$
Velocity in reversed-time is given by:
$$v_{rev}(t) = \frac{d}{dt}x_{rev}(t)$$
We can also express reversed-time velocity in terms of forward-time velocity:
$$\frac{d}{dt}x_{rev}(t) = \frac{d}{dt}x_{fwd}(-t) \tag6$$
The right-hand side of (6) is the rate of change in forward-time of position function given by:
$$x_{fwd}(-t) \tag{7}$$
It’s important to realize that the right-hand side of the equation (6) is not the velocity in forward-time, evaluated at time -t. Instead, it’s a forward-time rate of change of (7).
$$\frac{d}{dt}x_{fwd}(-t) \neq v_{fwd}(-t)$$
In order to express the right-hand of (6) in terms of forward-time velocity defined in (3) we need to use the chain rule for derivatives of composite functions. Let’s substitute:
$$u = -t$$
And apply a chain rule:
$$\frac{d}{dt}x_{fwd}(-t) = \left.\frac{dx_{fwd}(u)}{du}\right|_{u=-t}\left.\frac{du}{dt}\right|_{t} = -\left.\frac{dx_{fwd}(u)}{du}\right|_{u=-t} = -\frac{dx{fwd}}{dt}(-t) = -v_{fwd}(-t)$$
This means that reversed-time velocity evaluated at time t is equal to the negative of forward velocity evaluated at time -t:
$$v_{rev}(t) = -v_{fwd}(-t)$$
When we plug in -t to the original equation for force in forward-time (equation (4)), we get:
$$F_{fwd}(-t) = mv_{fwd}(-t)$$
After multiplying both sides by -1:
$$-F_{fwd}(-t) = -mv_{fwd}(-t)$$
When we replace forward time velocity with reverse time velocity, we have:
$$-F_{fwd}(-t) = mv_{rev}(t) \tag{8}$$
Equation (8) tells us that Aristotle’s law is deterministic into the past: reversed-time velocity at each point can be attributed to a specific value of a known force function.
Question #2: How comparing equations (1) and (2) would lead me to a conclusion that there is a “different rule for force as a function of time”?
Equation (5) describes Aristotle’s law of motion in reversed time:
$$F_{rev}(t) = mv_{rev}(t)$$
The right-hand sides of equations (5) and (8) are equal, which means that we can express reversed-time force in terms of the forward-time force.
$$F_{rev}(t) = -F_{fwd}(-t)$$
Now, back to the sentence that made me pause and think for a very long while:
The implication is simple: The reversed equation of motion is exactly like the original, but with a different rule for the force as a function of time.
If the rule for force is exactly the same in reversed-time, as it was in forward time, then I should have:
$$F_{rev}(t) \overset{?}{=} F_{fwd}(t) \tag{9}$$
Which means that the following would need to be true:
$$F_{fwd}(t) \overset{?}{=} -F_{fwd}(-t) \tag{10}$$
Does equation (10) automatically tell me that equation (9) is not true?
$$F_{rev}(t) \overset{?}{\neq} F_{fwd}(t)$$
No, since equation (10) is satisfied by any odd function. If I don’t know whether the force function is odd, I cannot conclude that a different force is needed in reversed-time for the Aristotle’s law to be applicable to reversed-time motion.
I can only arrive at the conclusion of a “different rule for force as a function of time”, if I already know that the force function is not odd in time.