### There are no zeros in physics

I recently read an article by Joseph Ford: “How random is a coin toss?” (1983). In it, Ford talks about the relation between completely deterministic systems and the seemingly random behaviour they sometimes produce. “Roulette wheel spins, dice throws […] are universally presumed to be completely random despite their obvious underlying determinism. Weather, human behavior and the stock market are, on the other hand, commonly regarded as strictly deterministic, notwithstanding their seemingly frivolous unpredictability.”

Through arguments on chaotic orbits and algorithmic complexity, he claims that eventually we must throw away our continuum descriptions of physics (this is regardless of quantum mechanics; see footnote), as those require calculations and measurements with infinite precision; however, as our computers are finite and our measurements imprecise, any actual calculation we do will have errors. In chaotic systems, which vastly outnumber the non-chaotic ones, these errors increase exponentially, and our predictions break down.

Instead of a continuum, we will use a number set that does not rely on infinite precision: only those numbers which can be algorithmically computed in finite time. This pretty much means only the rationals, it would seem. Now, in a way, we already roughly do this, and every time we run numerical simulations we restrict ourselves to the rationals, plus maybe a few other selected numbers. However, this is not included in our main physical theory, but is a byproduct of our finite machines.

An interesting point which especially caught my attention is:
“We thus eliminate the last infinities, the infinitely large

$\infty = (1+1+1+...)$

and the infinitely small

$0 = (1+1+1+...)^{-1}$

And indeed, while checking for equality is a difficult topic in floating point computations, our computers still allow us to assign the perfectly rational “0” to a variable. Yet in reality, it is very unlikely to encounter a physical property with a perfect 0 value. As a statistical, averaged property I might expect it to appear quite a lot, but as a deterministic one – less so.

Here are two nice points of wisdom I encountered during my studies which emphasize this:

1) I remember clearly one of our classes in Waves in the third semester. We were talking about damped harmonic motion, and the professor, analysing the damping coefficient, said something along the lines of: “we see here that there are two types of solutions: overdamping, if ξ > 1, and underdamping, if ξ < 1”. At this point, a student raised his hand and asked, “what about the critical damping, for ξ = 1?”. To this the professor replied: “nonsense, that is just a mathematical joke; a true physical system will never have a ξ value equal to exactly one!”

2) In solid state, and in practically any field in physics, whenever there is a local minimum in a potential, the physicists like to say that at that point, the system behaves as a harmonic oscillator. Why? Because then the derivative is 0, and the second derivative is positive. Looking at the Taylor expansion:

$V(x)=x_0+V'(x_0)(x-x_0)+\frac{1}{2}V''(x_0)(x-x_0)^2+O((x-x_0)^3)$

We know that $V'(x_0)=0$, and if we neglect the higher order terms, we get the equations for a harmonic oscillator. A question naturally arises: why must the second derivative be positive? Maybe it too is 0, and the smallest power in the Taylor series is at least 3? My Solid State TA gives the following answer: when dealing with physical systems, it’s so enormously hard to actually get a 0, that in essence the quadratic coefficient is almost always finite; the curvature must be very unique for both the first and second derivatives to be 0.
A good follow up question would then be: if it’s so hard to get a 0, why can we assume that the first derivative is 0? My answer to this: this stems from the fact that the potential has an actual extremum; having or not having an extremum is a physical binary quality, and so appears in physical variables. If we throw a ball up and it comes back down, we know that at some point it reached some maximal height and had a velocity 0 (even if we cannot pinpoint the exact moment of time with infinite precision, we know it exists and can compute around it, at least using continuum physics). A unique curvature, however, is in the continuum.

*Note: Ford mentions, but does not expand much on, the use of infinite precision in quantum mechanics, which is usually deterministic in its evolution but statistical in its measurements. I tend to see strength in his arguments only when we are purely deterministic.