Can the value of potentials be an irrational number?

Question

I was looking at the boundary conditions of Laplaces equation and I wondered if it is possible, given that the boundary conditions allows for a finite potential, that the potential can be an irrational value such as $e$ or some other form? It is possible to write some irrationals as a summation so I was wondering if a potential can be irrational? If so, do we say it's a finite solution?

Answer 1

I was looking at... Laplace's equation and I wondered if it is possible... that the potential can be an irrational value such as e or some other form?

Yes, it is possible.

...I was wondering if a potential can be irrational?

Yes, the value of the potential can be an irrational number.

Answer 2

In physics both discontinuous (e.g. the rectangular well) and continuous (e.g. the harmonic potential) potentials are used. A continuous potential that is not constant must change its value continuously from one value to another: this means that, in between those two values, it will assume infinitely many rational values and (even more) infinitely many irrational numbers.

The same is true for continuous field configurations (including boundary conditions), particle trajectories, the wavefunction in quantum mechanics, etc... practically every time you deal with ordinary or partial differential equations.

Answer 3

To specify whether any physical value is rational or irrational, you need to specify it infinitely precisely. There are infinitely many rational and irrational numbers between $2.718281828459045$ and $2.718281828459046$.

Classical physics is an approximation to the real world. Electrons are treated as point particles with no size. You can specify their position with as much precision as you like.

Given this, there is no reason you can't choose electrons separated by a distance that is an irrational number of meters.

To exactly calculate the potential energy of the two electron system, you need an exact value for the charge of an electron in Coulombs. It hasn't been measured exactly. But again you can approximate with the best measured value to as many digits as known, and use all $o$'s after that. You can calculate a number that is irrational in your approximation.

If you tried to make an infinitely precise measurement, you would find that the small difference between the real behavior of the universe and the classical approximation becomes important. An electron is better described by quantum mechanics. You can't specify the position and momentum with infinite accuracy, so you can't make an infinitely precise measurement. Likewise you can't make an infinitely precise measurement of energy in less than an infinite time.

So it doesn't really make sense to ask if the potential energy really is a rational or irrational number.

Answer 4

Let $V$ denote the potential at a given point. Your question is asking whether $V\in\mathbb{Q}$. There are a couple of problems with this. First, potentials are not dimensionless quantities, so where it rational or not depends on the choice of units, but we can address this by simply specifying a particular unit beforehand. Second, potentials are generally not gauge invariant, which means that you can $V$ to whatever real number you want, since it is only potential differences that actually matter. However, this does not answer the question of whether the potential difference between two points must be rational. Since field theories treat potentials as smooth (or at least continuous) functions on a smooth manifold, we can use the Intermediate Value Theorem to prove that a potential function must take on irrational values at some points, and a similar argument can be used for the potential difference between two points.

From the arguments above, we can see that the value of a potential not only can be, but at some points, must be irrational. This is a nice, rigourous, mathematical argument, but it also seems to be missing something, since after all, math isn't physics. In physics, we rely on empirical measurements that always have some nonzero uncertainty, and since the rationals and irrationals are both dense in the reals, there is no empirical way to distinguish between a rational and an irrational quantity. Thus, from an experimental perspective, it is effectively meaningless to ask whether the value of a potential is is irrational number.

Answer 5

There's no need to invoke strange physical scenarios or examine solutions to Laplace's equation, just take a look at the most elementary potential of all: a single point charge at $ \vec{r}' $ in vacuum, $$ V(\vec{r}) = \frac{1}{4 \pi \epsilon_0} \frac{Q}{|\vec{r} - \vec{r}'|} $$ Certainly you wouldn't claim that $ \pi $ or $ \epsilon $ are rational, right? With the exception of some specially selected values of $ \vec{r} $, the potential is irrational for almost every single point $ \vec{r} \in \mathbb{R}^3 $ because of these constants. Your question is entirely backward: rational potentials are the unthinkable result, not the other way around.

To go out on a limb, I suspect your question is intended to be a different philosophical one, where you meant to ask "can an irrational value of the potential be measured in the laboratory?", to which the answer is a resounding no. Irrational numbers are represented by an infinite and non-repeating sequence of decimals, but the measurement of such a number would require a piece of equipment capable of measuring the potential with infinite precision. You can't do that, though, and you will only be able to claim that a certain number of decimals in your measurement are reliably true. For example: with a ruler, you would never claim to be able to measure something in centimeters smaller than about $ 0.01 $cm---you would never, ever say your eyes can reliably measure $ 0.0138261826193... \, $cm. Your eyes (and all other measurement equipment in the universe) is limited as such. As a result, any value you measure will have a finite number of decimals, and appear to be rational.