Was Georg Simmel right in his mathematical argument against Nietzsche?

Question

As you can see in this link, the following argument originated from Georg Simmel was put forward against Nietzsche:

Even if there were exceedingly few things in a finite space in an infinite time, they would not have to repeat in the same configurations. Suppose there were three wheels of equal size, rotating on the same axis, one point marked on the circumference of each wheel, and these three points lined up in one straight line. If the second wheel rotated twice as fast as the first, and if the speed of the third wheel was 1/π of the speed of the first, the initial line-up would never recur.

I'm not here to ask about the philosophy, but rather the validity of Simmel's argument. Can't they really recur back to their original state even if it takes so much time?

My take: Even though one of the wheel's speed is irrational, it's still a constant, so I think we can pass his assumption of having an irrational number as a wheel's speed. Having that, shouldn't we be able to reach a point where they will end up back to their original state like many least common multiple problems do (although as I have found while trying to solve it, it's not as those problems)?

Answer 1

If you just look at the first wheel going 1 rotation per unit time, and the other wheel going $1/\pi$ rotations, they will never again line up (I do not know why that example has 3 wheels).

Let $r_1(t)$ and $r_2(t)$ be the radians spun by the first and other wheel over time $t\geq 0$. Then:

• $r_1(t) = 2\pi t$

• $r_2(t) = \frac{2\pi t}{\pi} = 2 t$

An integer number of rotations means an integer multiple of $2\pi$ radians. So to line up again at some time $T>0$, we need positive integers $n,k$ so that

\begin{align} \underbrace{r_1(T)}_{2\pi T}&= 2 \pi k\\ \underbrace{r_2(T)}_{2T}&= 2\pi n \end{align}

So we need $2 \pi T = 2 \pi k, 2 T = 2\pi n$, so $T = k = \pi n$, that is: $$ \pi = \frac{k}{n} $$ This is impossible since $\pi$ is irrational.

Answer 2

Atleast in the example given above George Simmel is wrong. Can u really give a wheel the speed of 1/pi? Since pi is an irrational eventually you'd have to truncate the (1/pi) number upto certain decimal point This will hence make the speed rationally representable. And because of that the wheels eventually will coincide. The main question is ,does there exist a physical process that can impart a non rational speed to an object.If answers is yes, then (only for this example) Simmel would be correct

Answer 3

Since this is Math SE, I’ll address the mathematical content of his question only. There are good questions to be asked about if this set up is physically possible or if it amounts to a serious philosophical challenge to Nietzsche, but those are not within the preview of this website (for what it’s worth, my answer to both are “no”).

First, let’s start with a simpler version:

Suppose three wheels are set up in the fashion described, one moving at a rate that we’ll normalize to $1$, one moving twice as fast, and one moving five times as fast. Will they ever line up?

WLOG, lets suppose the circumference of the wheel is a natural number, $n$. Let’s mark the wheel at unit distances around with $0,1,2,\ldots n-1$. The dots start at position $0$. The three dots will line up at their original positions eventually. This happens at time $t$ that satisfies $t=2t=5t\pmod{n}$. This has infinitely many solutions, and in particular has a solution whenever $t$ is a multiple of $n$. There may be others as well, depending on the prime factors of $n$.

Another way to think about this is to consider three lines, marked at every multiple of $n$. Take a point starting at $0$ and moving at the aforementioned rates. Now we wish to know if there will be a time where all three points reach a marked multiple of $n$ at the same time.

The advantage of this formulation is that it more easily accommodates transcendental velocities. This formulation of the problem (with the speeds that the OP specifies) will never have a solution. To see this, note that you would need there to be a $t$ and $k_1,k_2,k_3\in\mathbb Z$ such that $nk_1=t,nk_2=2t,nk_3=\frac{1}{\pi}t$ are all simultaneously satisfied. Combining the first and third gives us $nk_1=\pi nk_2$. Since $n,k_1,k_2$ are integers and $\pi$ is irrational, there is no solution.

So yes, the underlying mathematics of this objection is correct, independent of the physics or philosophical content.