How do we deal with irrationals in Physics? [duplicate]

This question already has an answer here:

This question is really very basic: how do we deal with irrationals in Physics? If for instance, in some meaningful calculation we get a length of $\pi$ meters, or a force of $\pi$ Newtons, how should we understand this? One could say: "well, we understand it as a rational approximation", but then there comes another question: which rational approximation? We can approximate $\pi$ by $3,14$ or $3,1415$ and so forth, increasing the precision, but yet never getting really to $\pi$.

Another problem is: well, the equations gave us $\pi$, and this is not a physically meaningful thing, so is there something wrong with the equation? Because, in nature I think there really isn't irrationals lurking arround, so that this becomes a little complicated.

So: how do we deal with irrationals in Physics?

marked as duplicate by Ben Crowell, Qmechanic♦Sep 10 '13 at 0:03

• – Qmechanic Sep 9 '13 at 22:54
• I vaguely remember that a famous physicist claimed that for whatever description of any physical entity in the universe less than 100 digits of $\pi$ are sufficient. See also physics.stackexchange.com/q/9621. – Alexander Sep 9 '13 at 22:59
• This question seems to be based on an unsupported (or should I say irrational? ::rimshot::) assumption that there is something wrong with irrational numbers. Where did you get that idea? – dmckee Sep 10 '13 at 2:45
• Well, I never thought they have something wrong, it just happens that some physicists I know said that they couldn't represent physical quantities and then I wanted to confirm how do we deal this in Physics. Now I see that who told me this is wrong and that irrationals can perfectly represent physical quantities. – user1620696 Sep 10 '13 at 12:32
• Are you sure what they told you was not you that complex numbers (i.e., "irreal") couldn't represent physical quantities? – leftaroundabout Sep 10 '13 at 13:03

You just do. If a calculation gives you $\pi$, then you just say it gave you $\pi$. It is a real number and it is as good as any rational in describing any numeric quantity: $$\text{all physical quantities are described by }\mathbb{REAL}\text{ numbers}.$$ If you must insist on what real numbers are and how we can even conceive of such a number as $\pi$, then you need to look carefully at your definition. As far as physics goes, the definition of the real numbers is

This is independent of the multiple possible realizations of this set of axioms from a "more basic" set, and of the multiple representations which you might think of using (such as decimal expansions, continued fractions, Dedekind cuts, equivalent classes of Cauchy sequences, or what-have-you). In describing numeric quantities, we need a structure that allows us to

• add, substract, multiply and divide as usual,
• compare one number to another and always conclude that one is greater than (or equal to) the other, and
• take limits, infima, suprema, derivatives, integrals, and so on.

The definition above embodies all these requirements. Since there is a unique such structure (up to a natural isomorphism), we don't really care what that structure is meant to be.

"But, but, but...", I hear you say, "what about rational approximations? We can never measure an irrational number!" And yes, that is correct. But we can't measure rational numbers, either. Our measurements come with a central value and a finite, nonzero precision: they describe an interval.* If our calculations come out to an irrational number, then all we can say is whether that number is consistent with experiment, and it's exactly the same with a rational number. Say that some calculation predicted a value of 4/3 for some quantity: then, in your own words,

we can approximate $1.33$, $1.3333$, and so forth, increasing the precision, but yet never getting really to $4/3$.

Such is life, and such is the mathematical structure of physics.

*or a probability distribution, if you insist. Same difference, though.

• We can approximate 1 by 1.0, 1.00, 1.000, 1.0000, and so forth! The only difference is these are all zeros, in 4/3 = 1.33333... these are all threes, in the case of Pi... well, some random-looking damn digits. So? In any case, in any measurement, your precision is limited. So your 1.0m stick is only good to 100 microns anyway. 3.14159 is good enough in that case. – safkan May 5 '16 at 20:26

the equations gave us $\pi$, and this is not a physically meaningful thing...

What is "a physically meaningful thing" then? I suppose anybody would safely say that natural numbers are physically meaningful, you can certainly count separate entities.

But the trouble starts much earlier than irrational: what's a negative number supposed to be? I don't think you can give a sound physical definition, you need to resort to maths. Similarly for $\mathbb{Q}$: you can define a positive rational number as a tuple of natural numbers and describe how to add/multiple/divide those, but ab initio there's not much physical about this.

Incidentally, the completion to $\mathbb{R}$ is where things become physical again: unlike the rationals, the reals are a continuum. The usual mathematical definition goes via Cauchy sequences, which is actually more physical than one might think: it's a lot like measuring some quantity with ever-increasing precision. You get value after value in some sequence, and each time there's some uncertainty. But if you can limit that uncertainty for each measurement, and are able to make it go to zero, then the sequence of measurements converges towards the "true physical value" of the measured quantity. (Of course, this is never really possible due to quantum limitations.)

It boils down to this: maths by itself is never physically meaningful, you can only give it a physical meaning by mapping mathematical objects to measurement outcomes in some way. That rational numbers are in any way special and "more pure" is an idea going back to the Pythagoreans. From a modern point of view there aren't really good arguments to back this.