A few weeks ago I started wondering if there's any reason for a given fundamental physical constant, like Boltzmann's one, to enjoy some algebraic properties, for example being rational/irrational or algebraic/transcendental, in a given system of units, of course. Note that, once every fundamental constant is a real number, every fundamental constant is algebraic over $\mathbb{R}[x]$: you can take the polynomial $x-a\in\mathbb{R}[x]$, where $a$ is the given fundamental constant. So the algebraic/transcendental mentioned before is over $\mathbb{Q}[x]$.

I know that, given a fundamental constant $a$ you can consider a system of units in which $a=1$. But the question is, given a system of units, MKS for instance, what can we say about the fundamental constants at that system of units: are they rational numbers? If not, are they algebraic numbers (over $\mathbb{Q}[x]$)?

I know that maybe this question is of no importance to physics (nature doesn't care if Planck's constant is rational or not), but I was just curious. If you guys don't think so, feel free to explain why.

  • 1
    $\begingroup$ Physics does not make a difference between algebraic and transcendental numbers because any two such numbers which differ only by an amount that is smaller than the measurement errors of the best experiments determining them are in the equivalence class of all numbers that satisfy the hypothesis that they parametrize. $1$ and $1 + 10^{-20}*\pi$ are identical for physicists in all scenarios I am aware of. $\endgroup$
    – CuriousOne
    Jun 12, 2015 at 6:00

1 Answer 1


Since most constants aren't calculated theoretically, but rather they are measured experimentally, this question is impossible to work out. (Or, if calculated theoretically, one uses measured constants for the calculation)

Unless you reach infinite precision (Which is impossible in more than one aspect), there's no difference in measuring a rational or irrational constant. Moreover, as you've said, it's quite arbitrary anyway, since our system of units is arbitrary.

  • $\begingroup$ it would be great if the fine structure constant turns out to be $1/137$ for some strange theoretical reason. (it's an adimensional constant) $\endgroup$
    – Héctor
    Jun 12, 2015 at 13:27
  • 1
    $\begingroup$ See en.wikipedia.org/wiki/Fine-structure_constant#Measurement - It's not exactly that, there are loop corrections, and it's one of the best measured physical constants. $\endgroup$
    – Omry
    Jun 12, 2015 at 14:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.