Standard treatments of the Buckingham Pi Theorem seem to imply that given a dimensionless function $f$ of variables $q_1, q_2, \dots, q_n$ with associated dimension matrix having rank $r$, there exists a function $\phi$ of $\nu = n-r$ variables and pi groups $\pi_1, \pi_2, \dots, \pi_\nu$ such that $$ f(q_1, q_2, \dots, q_n) = \phi(\pi_1, \pi_2, \dots, \pi_\nu). \tag{$\star$} $$ A pi group is a function of the $q_i$ that takes the form $q_1^{a_1}q_2^{a_2}\cdots q_n^{a_n}$ for some rational $a_1, a_2, \dots, a_n$. If all the $a_i$ are zero, I'll call the pi group "trivial."

A simple counterexample

Consider the case $n=1$ with $q_1 = q$ for notational simplicity, and let the function $f:\mathbb R\to \mathbb R$ be defined by $$ f(q) = \mathrm{sgn}(q) $$ where $\mathrm{sgn}$ is the sign function. This function has the property that $$ f(\lambda q) = f(q) $$ for all positive $\lambda$ -- it's scale-invariant under positive changes of scale, so no matter what dimensions one assigns to $q$, this function is dimensionless, but it cannot be written as a function of pi groups. Indeed if $q$ is dimensionful, then there will be no non-trivial pi groups to speak of, and $\mathrm{sgn}(q)$ is not a pi group.

A less reductive/contrived but essentially equivalent counterexample

Consider the formula for quadratic drag in one dimension: $$ F_d(\rho, A, v) = -\frac{1}{2}C_d\rho A v|v| = -\frac{1}{2}C_d\rho A v^2\mathrm{sgn}(v). $$ Dividing both sides by $\rho A v^2$ reveals a dimensionless quantity $$ f(\rho, A, v) = \frac{F_d(\rho,A, v)}{\rho A v^2} = -\frac{1}{2}C_d\mathrm{sgn}(v) $$ which cannot be written in terms of pi groups. There are no non-trivial pi groups that can be formed from density, area, and velocity, and $\mathrm{sgn}(v)$ is not a pi group.

Error in proof?

In addition to these counterexamples, as I was reading a proof of the Pi Theorem in Bluman and Kumei's Symmetries and Differential Equations, I came across a step in the proof (the statement "Consequently $F$ is independent of $X_n$" right after eq. 1.23) that I believe is erroneous given that we only allow for positive scale transformations when changing units, and I think this error corresponds to missing the possibility of needing both pi groups and sign functions of variables in writing dimensionless quantities in terms of other dimensionless quantities. The crux of the problem with the argument seems to me the assertion that if a function is positive scale invariant, namely $f(\lambda q) = f(q)$ for all positive $\lambda$, then it is constant. The sign function is a counterexample to this assertion. I think the most that can be said about a function $f:\mathbb R\to\mathbb R$ that is positive scale invariant is that it's constant for all negative values and constant for all positive values for potentially different constants.

My question in a nutshell

If a dimensionless quantity $f$ is a function of non-negative quantities $q_1, q_2, \dots, q_n$, then it seems the standard statement of the theorem is correct -- e.g. the proof in Bluman and Kumei goes through without error as far as I can tell, but if one or more variables is allowed to be negative, shouldn't equation ($\star$) above instead be written as $$ f(q_1, q_2, \dots, q_n) = \phi(\mathrm{sgn}(q_1), \mathrm{sgn}(q_2), \dots, \mathrm{sgn}(q_n), \pi_1, \pi_2, \dots, \pi_\nu)? \tag{$\star\star$} $$ I believe that Bluman and Kumei's proof can be modified to prove this slightly weaker version of the theorem.

  • 2
    $\begingroup$ Leaving aside the issue of the $\text{sgn}$ function, is your example a correct application of the theorem? Shouldn't your dimensionless function be $f(\rho, A, v, F_d)$ not $f(\rho, A, v)$? You then do have a dimensionless group, namely $F_d/(\rho A v^2)$. The fact that this happens to be a discontinuous function involving $\text{sgn}(v)$ is irrelevant. $\endgroup$
    – alephzero
    Jan 2, 2017 at 8:29
  • $\begingroup$ @alephzero If the "dimensionless groups" referred to in the statement of the theorem are allowed to contain e.g. $\mathrm{sgn}(v)$, then I have no objection(s). However, I don't think this is so in the versions and "proofs" I have seen. See, for example, the Wikipedia page statement in which a "pi group" is defined to have the form $q_1^{a_1}\cdots q_n^{a_n}$ for some rationals $a_1, \dots, a_n$. $\endgroup$ Jan 2, 2017 at 8:38
  • $\begingroup$ Does the problem still occur if you replace the sign function with a smooth interpolating function, which does the interpolation over a dimensionful, physical scale $v_0$? $\endgroup$
    – knzhou
    Jan 2, 2017 at 23:40
  • $\begingroup$ @knzhou That's an interesting idea. If, in addition to each variable $q_i$ one has access to a corresponding small, positive dimensionful scale $q_{i,0}$ with the same dimensions, then replacing each $\mathrm{sgn}(q_i)$ with $\mathrm{sigmoid}(q_i/q_{i,0})$ for some sigmoid-type function would, I believe, allow one to use only pi groups as in the standard theorem, but then you'd have to introduce extra scales which seems a higher price to pay than just using $\mathrm{sgn}$ no? $\endgroup$ Jan 3, 2017 at 1:24
  • 2
    $\begingroup$ @joshphysics I wouldn't call it an extra scale, though. Physically, there has to be some scale like this in the problem, because the sign function doesn't actually occur in nature. So it's more like, the original example looks like it violates the theorem because it ignores the small velocity scale! $\endgroup$
    – knzhou
    Jan 3, 2017 at 1:31

1 Answer 1


In application of Pi theorem (the way I have applied it) we only seek relationship between magnitudes of quantities involved in the problem. Vector character of any of the quantities does not enter the picture. In writing $F_D=C_D\frac{1}{2}\rho A v^2~sgn(v)$, you have brought in the vector character of force into the equation, and in itself this is correct. However Pi theorem relates only scalar magnitudes, and demanding that it account for the sign of physical quantities as well is to ask too much of it. In particular Pi theorem gives the dimensionless groups that would remain invariant under change of measurement units, which are always positive (hence the restriction to positive scale transformations).

  • $\begingroup$ If it's really the case that only magnitudes are being considered in standard treatments of the theorem, then I'm satisfied, but is that ever made explicit? (references?) Is asking for the theorem to account for the signs of physical quantities really too much? By modifying the standard proof only slightly, one obtains the version $(\star\star)$ that accounts for signs. I agree that positive scale transformations are appropriate. It's the way that positive scale transformations seem to be used in B&K's proof that I object to. This objection disappears if the $q_i$ are all positive. $\endgroup$ Jan 3, 2017 at 19:20
  • $\begingroup$ Indeed your version (**) is correct if you wanted to include sign dependence. But is it more useful? If you think as a mathematician, then you don't worry about its usefulness and just go in for the most general treatment of the problem. But I who do experiments, would like to have as few dimensionless groups as possible. It is difficult enough to investigate a relationship involving even three dimensionless groups (excluding sign dependence). In summary, I admit that you are right in the interest of generality, but in applications that degree of generality is unhelpful. $\endgroup$
    – Deep
    Jan 4, 2017 at 4:37
  • $\begingroup$ That's reasonable, thanks for your thoughts. I realize that this question doesn't have enormous practical appeal, but since the theorem is (in many versions I have read -- many by mathematicians) presented as having purely mathematical content, I was surprised that there seems to be a bit of sloppiness on this point. $\endgroup$ Jan 4, 2017 at 6:01

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